Solve in real numbers:
1
x - ⌊x⌋ = -----------
1/x + 1/⌊x⌋
Note:⌊x⌋ is the floor of x, which is the greatest integer less than or equal to x.
I'm sure there is a shorter way but here goes:
First, is x=0 a solution?
LHS = 0
RHS = (x*⌊x⌋ / (x + ⌊x⌋) = 0/0 (undefined)
So, no, x=0 is not a solution
Next, can x be negative?
No, because LHS is always a positive fraction but the RHS would be one over the sum of two negative fractions.
Let x = I + f
⌊x⌋ = I
1
x - ⌊x⌋ = ------------
1/x + 1/⌊x⌋
x - ⌊x⌋ = f
f/x + f/⌊x⌋ = 1
f/x + f/I = 1
If + fx = Ix
substitute x = I+f
If + f(I+f) = I(I+f)
2If + f^2 = I^2 + If
If + f^2 = I^2
I^2 - If - f^2 = 0
I = f(1 ± √5)/2
f^2 + If - I^2 = 0
f = I(-1 ± √5)/2 since f>0, only the + is valid
f = I(√5 - 1)/2, since f<1,
I < 2/(√5 - 1) +~ 1.618 = φ
So I=1 is the only solution for I, except I=0
If I=1, then f = (√5 - 1)/2 +~ 0.618
x = (1+√5)/2 = φ
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Posted by Larry
on 2024-10-14 10:04:46 |
Solution